B样条数据点反求控制点绘制曲线（源码）
一、軟件功能需求
1）所設計的軟件應具有圖形化用戶界面（GUI）；
2）用戶在軟件界面上可用隨機數方式或手工方式輸入若干曲線或曲面的數據點，例如起點、終點、列表型值點等，對于曲線，還可設置步長參數；對于曲面，還可設置步長與行距參數；曲線或曲面的類型不限。
3）具有“繪制理想圖形”按鈕，用戶完成數據點與參數輸入后，點擊該按鈕，軟件可繪制出理想的曲線或曲面，若該曲線或曲面有特征多邊形，則還能通過型值點反算出控制頂點，并繪制出對應的特征多邊形；
4）具有“生成加工路徑”按鈕，點擊該按鈕，可根據設定的步長與行距參數自動生成加工該曲線或曲面的刀具路徑；
5）具有“生成加工程序”按鈕，點擊該按鈕，可自動生成一個實現該曲線或曲面加工的G代碼數控加工程序文件。
二、B樣條反求控制點原理
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參考書籍：計算機輔助幾何設計與非均勻有理B樣條（一般學校的電子圖書館可以在線閱讀）
P212頁：B樣條遞推定義
P237頁：齊次B樣條曲線的節點矢量
P257頁：反算三次B樣條插值曲線的控制頂點
P259頁：邊界條件：拋物線條件
二、軟件實現
軟件平臺：Windows7?? Qt5.9.3???? 語言：C++
??? 由于默認界面框的原點在左上角，并且向下為Y+，向右為x+，所以在畫圖或者鼠標捕捉時，要將其轉化到正常的直角坐標系，并繪制坐標系。
??? QPainter* painter = new QPainter(this);
??????? painter->setWindow(-width()/2,height()/2,width(),-height());? //設置坐標原點在屏幕的正中心
??? ?
??? //??? painter->drawLine(-296,0,296,0);
??? //??? painter->drawLine(0,296,0,-296);
??????? //繪制x軸
??????? QPointF startPoint(-295,0);
??????? QPointF endPoint(295,0);
??????? double x1, y1, x2, y2; //箭頭的兩點坐標
??????? //求得箭頭兩點坐標
??????? calcVertexes(startPoint.x(), startPoint.y(), endPoint.x(), endPoint.y(), x1, y1, x2, y2);
??? ?
??????? painter->drawLine(startPoint, endPoint);//繪制線段
??????? painter->drawLine(endPoint.x(), endPoint.y(), x1, y1);//繪制箭頭一半
??????? painter->drawLine(endPoint.x(), endPoint.y(), x2, y2);//繪制箭頭另一半
??? ?
??????? //繪制Y軸
??????? startPoint.setX(0);
??????? startPoint.setY(-295);
??????? endPoint.setX(0);
??????? endPoint.setY(295);
??????? //求得箭頭兩點坐標
??????? calcVertexes(startPoint.x(), startPoint.y(), endPoint.x(), endPoint.y(), x1, y1, x2, y2);
??? ?
??????? painter->drawLine(startPoint, endPoint);//繪制線段
??????? painter->drawLine(endPoint.x(), endPoint.y(), x1, y1);//繪制箭頭一半
??????? painter->drawLine(endPoint.x(), endPoint.y(), x2, y2);//繪制箭頭另一半
??? ?
??? void Palette::calcVertexes(double start_x, double start_y, double end_x, double end_y, double& x1, double& y1, double& x2, double& y2)
??? {
??? double arrow_lenght_ = 10;//箭頭長度，一般固定
??? double arrow_degrees_ = 0.5;//箭頭角度，一般固定
??? ?
??? double angle = atan2(end_y - start_y, end_x - start_x) + 3.1415926;//
??? ?
??? x1 = end_x + arrow_lenght_ * cos(angle - arrow_degrees_);//求得箭頭點1坐標
??? y1 = end_y + arrow_lenght_ * sin(angle - arrow_degrees_);
??? x2 = end_x + arrow_lenght_ * cos(angle + arrow_degrees_);//求得箭頭點2坐標
??? y2 = end_y + arrow_lenght_ * sin(angle + arrow_degrees_);
??? }
??? 采用隨機數生成8個隨機點，對第一組隨機數進行排序，第二組隨機數不排序，填充到QTableWidget。
??? void MainWindow::on_general_random_clicked()
??? {
??????? QVector <int> a;??? //定義一個vector容器
??????? srand((int)time(NULL));? //時間作為種子
??????? for (int i=0;i<8;i++)
??????? {
??????????? a.append(rand()%296-148);? //向vector后面追加元素，區間根據顯示區域設定
??????? }
??????? qSort(a.begin(),a.end());?? //對vector從小到大排序
??? ?
??????? for(int i=0;i<8;i++) //向X坐標寫入數據
??????? {
??????????? ui->model->setItem(i,0,new QTableWidgetItem(QString("%1").arg(a[i])));
??????? }
??????? a.remove(0,8);? //刪去第一列
??? ?
??????? srand((int)time(NULL));
??????? for (int i=0;i<8;i++)
??????? {
??????????? a.append(rand()%296-148);
??????? }
??????? for(int i=0;i<8;i++)
??????? {
??????????? ui->model->setItem(i,1,new QTableWidgetItem(QString("%1").arg(a[i])));
??????????? ui->model->setItem(i,2,new QTableWidgetItem(QString("%1").arg(0)));
??????? }
??? }
?
??? 獲取數據點，根據數據點反求控制點。
??? void MainWindow::on_general_curve_clicked()
??? {
??????? QPoint point;
??????? QPoint start,end;
??????? QString QTW_data1,QTW_data2;
??????? double a1,a2;
??????? //提取數據表中數據
????????? for(int i=0; i<8; i++)
????????? {
????????????? QTW_data1=ui->model->item(i, 0)->text();
????????????? QTW_data2=ui->model->item(i, 1)->text();
??? ?
????????????? a1=QTW_data1.toDouble();
????????????? a2=QTW_data2.toDouble();
??? ?
????????????? point.setX(a1);
????????????? point.setY(a2);
????????????? if(i==0)
????????????? {
????????????????? start.setX(a1);
????????????????? start.setY(a2);
????????????? }
????????????? if(i==7)
????????????? {
????????????????? end.setX(a1);
????????????????? end.setY(a2);
????????????? }
??? ?
????????????? palette->typepoints.append(point);??? ?
??? ?
????????? }
????????? /*
??????????? 此段程序將數據點反求出控制點,八個數據點
????????? */
????????? double l[7];? //存放數據點之間的距離，l[0]代表第一個數據點到第二個數據點之間的距離
????????? double L=0;//存放總長度
????????? for(int i=0;i<7;i++)
????????? {
????????????? l[i]=sqrt(pow((palette->typepoints[i+1].x()-palette->typepoints[i].x()),2)+
????????????????????? pow((palette->typepoints[i+1].y()-palette->typepoints[i].y()),2)); //兩點之間距離公式
??? ?
?????????????? L=L+l[i];
????????? }
????????? //節點矢量計算
????????? for(int i=0;i<4;i++)
????????? {
????????????? palette->u[i]=0;
????????? }
????????? for(int i=0;i<4;i++)
????????? {
????????????? palette->u[i+10]=1;
????????? }
????????? palette->u[4]=l[0]/L;
????????? palette->u[5]=(l[0]+l[1])/L;
????????? palette->u[6]=(l[0]+l[1]+l[2])/L;
????????? palette->u[7]=(l[0]+l[1]+l[2]+l[3])/L;
????????? palette->u[8]=(l[0]+l[1]+l[2]+l[3]+l[4])/L;
????????? palette->u[9]=(l[0]+l[1]+l[2]+l[3]+l[4]+l[5])/L;
????????? for(int i=0;i<14;i++)
????????? {
????????????? qDebug()<<"u[i]="<<palette->u[i]<<endl;
????????? }
??? ?
????????? double detal[13];
????????? for(int i=0;i<13;i++)
????????? {
????????????? detal[i]=palette->u[i+1]-palette->u[i];
?????????????? qDebug()<<"detal[i]="<<detal[i]<<endl;
????????? }
????????? //根據節點矢量，反算控制點，其中需要使用矩陣求逆
??? ?
????????? double a[8],b[8],c[8],e[8],f[8];? //角標0-8
????????? a[0]=1-(detal[3]*detal[4])/pow((detal[3]+detal[4]),2);
????????? b[0]=(detal[3]/(detal[3]+detal[4]))*(detal[4]/(detal[3]+detal[4])-detal[3]/(detal[3]+detal[4]+detal[5]));
????????? c[0]=(pow((detal[3]),2))/((detal[3]+detal[4])*(detal[3]+detal[4]+detal[5]));
????????? e[0]=(1.0/3)*(palette->typepoints[0].x()+2*palette->typepoints[1].x());? //X軸坐標
????????? f[0]=(1.0/3)*(palette->typepoints[0].y()+2*palette->typepoints[1].y());? //Y軸坐標
??? ?
??? ?
????????? a[7]=-(pow((detal[9]),2))/((detal[8]+detal[9])*(detal[8]+detal[8]+detal[9]));
????????? b[7]=(detal[9]/(detal[8]+detal[9]))*(detal[9]/(detal[8]+detal[8]+detal[9])-detal[8]/(detal[8]+detal[9]));
????????? c[7]=detal[8]*detal[9]/pow((detal[8]+detal[9]),2)-1;
????????? e[7]=-(1.0/3)*(palette->typepoints[7].x()+2*palette->typepoints[6].x());
????????? f[7]=-(1.0/3)*(palette->typepoints[7].y()+2*palette->typepoints[6].y());
??? ?
??? ?
????????? for(int i=1;i<7;i++)
????????? {
???????????? // e[8]=palette->typepoints[i].x();
????????????? a[i]=(pow((detal[i+3]),2))/(detal[i+1]+detal[i+2]+detal[i+3]);
????????????? b[i]=detal[i+3]*(detal[i+1]+detal[i+2])/(detal[i+1]+detal[i+2]+detal[i+3])+detal[i+2]*(detal[i+3]+detal[i+4])/(detal[i+2]+detal[i+3]+detal[i+4]);
????????????? c[i]=pow((detal[i+2]),2)/(detal[i+2]+detal[i+3]+detal[i+4]);
????????????? e[i]=(detal[i+2]+detal[i+3])*palette->typepoints[i].x();
????????????? f[i]=(detal[i+2]+detal[i+3])*palette->typepoints[i].y();
????????? }
??? ?
????????? for(int i=0;i<8;i++)
????????? {
????????????? qDebug()<<"a[i]="<<a[i]<<"b[i]="<<b[i]<<"c[i]="<<c[i]<<"e[i]"<<e[i]<<endl;
????????? }
??? ?
????????? double matrix[8][8];
????????? for(int i=0;i<8;i++)
????????? {
????????????? for(int j=0;j<8;j++)
????????????? {
????????????????? matrix[i][j]=0;
????????????????? qDebug()<<matrix[i][j]<<" ";
????????????? }
????????????? qDebug()<<endl;
????????? }
??? ?
????????? qDebug()<<"_________________________"<<endl;
??? ?
????????? /*對系數矩陣賦值*/
????????? matrix[0][0]=a[0];
????????? matrix[0][1]=b[0];
????????? matrix[0][2]=c[0];
??? ?
????????? for(int i=1;i<7;i++)
????????? {
????????????? matrix[i][i-1]=a[i];
????????????? matrix[i][i]=b[i];
????????????? matrix[i][i+1]=c[i];
????????? }
??? ?
?????????? matrix[7][5]=a[7];
?????????? matrix[7][6]=b[7];
?????????? matrix[7][7]=c[7];
??? ?
?????????? for(int i=0;i<8;i++)
?????????? {
?????????????? for(int j=0;j<8;j++)
?????????????? {
?????????????????? qDebug()<<matrix[i][j]<<" ";
?????????????? }
??? ?
?????????? }
?????????? qDebug()<<endl;
??? ?
?????????? double matrix_after[N][N];
?????????? qDebug()<<"_________________"<<endl;
??? ?
?????????? matrix_inv(matrix,matrix_after);? //求逆矩陣
?????????? for(int i=0;i<8;i++)
?????????? {
?????????????? for(int j=0;j<8;j++)
?????????????? {
?????????????????? qDebug()<<matrix_after[i][j]<<" ";
?????????????? }
??? ?
?????????? }
?????????? qDebug()<<endl;
??? ?
?????????? double sum_x[8]={},sum_y[8]={};
??? ?
?????????? palette->ctrlPoints.append(start);
?????????? for(int i=0;i<8;i++)
?????????? {
?????????????? for(int j=0;j<8;j++)
?????????????? {
?????????????????? sum_x[i]=sum_x[i]+matrix_after[i][j]*e[j];
?????????????????? sum_y[i]=sum_y[i]+matrix_after[i][j]*f[j];
?????????????? }
?????????????? palette->ctrlPoints.append(QPoint(sum_x[i],sum_y[i]) );
?????????????? qDebug()<<"sum_x[i]="<<sum_x[i]<<"sum_y[i]="<<sum_y[i]<<endl;
?????????? }
?????????? palette->ctrlPoints.append(end);
?????????? qDebug()<<palette->ctrlPoints<<endl;
??????? palette->generateCurve();? //根據控制點繪制型值點
??????? palette->update();???????? //更新畫板
??? }
?
??? 矩陣求逆C++實現：
??? void MainWindow::matrix_inv(double matrix_before[N][N],double matrix_after[N][N])
??? {
??????? bool flag;//標志位，如果行列式為0，則結束程序
??? ?
??????? flag=GetMatrixInverse(matrix_before,N,matrix_after);
??? ?
??? }
??? ?
??? //按第一行展開計算|A|
??? double MainWindow::getA(double arcs[N][N],int n)
??? {
??????? if(n==1)
??????? {
??????????? return arcs[0][0];
??????? }
??????? double ans = 0;
??????? double temp[N][N]={0.0};
??????? int i,j,k;
??????? for(i=0;i<n;i++)
??????? {
??????????? for(j=0;j<n-1;j++)
??????????? {
??????????????? for(k=0;k<n-1;k++)
??????????????? {
??????????????????? temp[j][k] = arcs[j+1][(k>=i)?k+1:k];
??? ?
??????????????? }
??????????? }
??????????? double t = getA(temp,n-1);
??????????? if(i%2==0)
??????????? {
??????????????? ans += arcs[0][i]*t;
??????????? }
??????????? else
??????????? {
??????????????? ans -=? arcs[0][i]*t;
??????????? }
??????? }
??????? return ans;
??? }
??? ?
??? //計算每一行每一列的每個元素所對應的余子式，組成A*
??? void? MainWindow::getAStart(double arcs[N][N],int n,double ans[N][N])
??? {
??????? if(n==1)
??????? {
??????????? ans[0][0] = 1;
??????????? return;
??????? }
??????? int i,j,k,t;
??????? double temp[N][N];
??????? for(i=0;i<n;i++)
??????? {
??????????? for(j=0;j<n;j++)
??????????? {
??????????????? for(k=0;k<n-1;k++)
??????????????? {
??????????????????? for(t=0;t<n-1;t++)
??????????????????? {
??????????????????????? temp[k][t] = arcs[k>=i?k+1:k][t>=j?t+1:t];
??????????????????? }
??????????????? }
??? ?
??? ?
??????????????? ans[j][i]? =? getA(temp,n-1);? //此處順便進行了轉置
??????????????? if((i+j)%2 == 1)
??????????????? {
??????????????????? ans[j][i] = - ans[j][i];
??????????????? }
??????????? }
??????? }
??? }
??? ?
??? //得到給定矩陣src的逆矩陣保存到des中。
??? bool MainWindow::GetMatrixInverse(double src[N][N],int n,double des[N][N])
??? {
??????? double flag=getA(src,n);
??????? double t[N][N];
??????? if(0==flag)
??????? {
??? //??????? cout<< "原矩陣行列式為0，無法求逆。請重新運行" <<endl;
??????????? return false;//如果算出矩陣的行列式為0，則不往下進行
??????? }
??????? else
??????? {
??????????? getAStart(src,n,t);
??????????? for(int i=0;i<n;i++)
??????????? {
??????????????? for(int j=0;j<n;j++)
??????????????? {
??????????????????? des[i][j]=t[i][j]/flag;
??????????????? }
??? ?
??????????? }
??????? }
??? ?
??????? return true;
??? }
??? 根據控制點繪制B樣條曲線：
??? void Palette::generateCurve()
??? {
??????? curvePoints.clear();
??????? int i=3;
??????? for(double u1=0; u1<1; u1+=0.001)
??????? {
??????????? QPointF tmp(0,0);
??????????? if(u1>u[i+1])
??????????? {i=i+1;}
??????????? for(int k=0; k<4;k++)
??????????? {
??????????????? QPointF t = ctrlPoints[i-k];
??? ?
??????????????? t*=Nu(i-k,u1,3);
??? ?
??????????????? tmp+=t;
??????????? }
??????????? //qDebug()<<"tmp="<<tmp;
??????????? curvePoints.push_back(tmp);
??????? }
??? }
??? ?
??? double Palette::Nu(int i,double u2,int k)
??? {
??????? if(k==0)
??????? {
??????????? if((u2>=u[i])&&(u2<u[i+1]))
??????????????? return 1;
??????????? else
??????????????? return 0;
??????? }
??????? else{
??????????? double a1=(u2-u[i]),c1=(u[i+k]-u[i]);
??????????? double b1=(u[i+k+1]-u2),d1=(u[i+k+1]-u[i+1]);
??????????? if(d1<0.001)
??????????? {
??????????????? b1=0;
??????????????? d1=1;
??????????? }
??????????? if(c1<0.001)
??????????? {
??????????????? a1=0;
??????????????? c1=1;
??????????? }
??????????? double nu=(a1/c1)*Nu(i,u2,k-1)+(b1/d1)*Nu(i+1,u2,k-1);
??????????? return nu;
??? ?
??????? }
??? ?
??? }
??? 運行結果
?
致謝：強強（B樣條原理）、瑤哥（矩陣求逆，遞推公式繪圖）
源碼：https://download.csdn.net/download/qq_32059343/10927125
————————————————
版權聲明：本文為CSDN博主「東辰葉落」的原創文章，遵循 CC 4.0 BY-SA 版權協議，轉載請附上原文出處鏈接及本聲明。
原文鏈接：https://blog.csdn.net/qq_32059343/article/details/86408359
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